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        <h3 id="一、题目描述"><a href="#一、题目描述" class="headerlink" title="一、题目描述"></a>一、题目描述</h3><p>&ensp;&ensp;&ensp;&ensp;反转一个单链表。</p>
<span id="more"></span>

<p>示例:</p>
<pre><code>输入: 1-&gt;2-&gt;3-&gt;4-&gt;5-&gt;NULL
输出: 5-&gt;4-&gt;3-&gt;2-&gt;1-&gt;NULL
</code></pre>
<p>进阶:<br>&ensp;&ensp;&ensp;&ensp;你可以迭代或递归地反转链表。</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">/**</span></span><br><span class="line"><span class="comment"> * Definition for singly-linked list.</span></span><br><span class="line"><span class="comment"> * struct ListNode &#123;</span></span><br><span class="line"><span class="comment"> *     int val;</span></span><br><span class="line"><span class="comment"> *     ListNode *next;</span></span><br><span class="line"><span class="comment"> *     ListNode(int x) : val(x), next(NULL) &#123;&#125;</span></span><br><span class="line"><span class="comment"> * &#125;;</span></span><br><span class="line"><span class="comment"> */</span></span><br><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> &#123;</span></span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">ListNode* <span class="title">reverseList</span><span class="params">(ListNode* head)</span> </span>&#123;</span><br><span class="line">		<span class="comment">//to do</span></span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>
<h3 id="二、解题思路"><a href="#二、解题思路" class="headerlink" title="二、解题思路"></a>二、解题思路</h3><p>&ensp;&ensp;&ensp;&ensp;当链表为空或者只有一个头结点时，直接返回head即可。<br>&ensp;&ensp;&ensp;&ensp;定义三个指针：</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">ListNode* p=head;</span><br><span class="line">ListNode* q=head-&gt;next;</span><br><span class="line">ListNode* tmp;</span><br></pre></td></tr></table></figure>
<p>&ensp;&ensp;&ensp;&ensp;p是反转链表的头节点，q是老链表的头节点，tmp作为p，q移动时的中间节点。<br><img src="/myblog/2021/04/30/%E9%93%BE%E8%A1%A8%EF%BC%9A%E5%8F%8D%E8%BD%AC%E9%93%BE%E8%A1%A8/1.png" alt="在这里插入图片描述"><br>&ensp;&ensp;&ensp;&ensp;移动一次</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br></pre></td><td class="code"><pre><span class="line">tmp=q;</span><br><span class="line">q=q-&gt;next;</span><br><span class="line">tmp-&gt;next=p;</span><br><span class="line">p=tmp;</span><br></pre></td></tr></table></figure>

<h3 id="三、我的代码"><a href="#三、我的代码" class="headerlink" title="三、我的代码"></a>三、我的代码</h3><p>&ensp;&ensp;&ensp;&ensp;直接在力扣上写的。</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">/**</span></span><br><span class="line"><span class="comment"> * Definition for singly-linked list.</span></span><br><span class="line"><span class="comment"> * struct ListNode &#123;</span></span><br><span class="line"><span class="comment"> *     int val;</span></span><br><span class="line"><span class="comment"> *     ListNode *next;</span></span><br><span class="line"><span class="comment"> *     ListNode(int x) : val(x), next(NULL) &#123;&#125;</span></span><br><span class="line"><span class="comment"> * &#125;;</span></span><br><span class="line"><span class="comment"> */</span></span><br><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> &#123;</span></span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">ListNode* <span class="title">reverseList</span><span class="params">(ListNode* head)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">if</span>(head==<span class="literal">nullptr</span>||head-&gt;next==<span class="literal">nullptr</span>)</span><br><span class="line">            <span class="keyword">return</span> head;</span><br><span class="line">        ListNode* p=head;</span><br><span class="line">        ListNode* q=head-&gt;next;</span><br><span class="line">        ListNode* tmp;</span><br><span class="line">        p-&gt;next=<span class="literal">nullptr</span>;</span><br><span class="line">        <span class="keyword">while</span>(q!=<span class="literal">nullptr</span>)&#123;            </span><br><span class="line">            tmp=q;</span><br><span class="line">            q=q-&gt;next;</span><br><span class="line">            tmp-&gt;next=p;</span><br><span class="line">            p=tmp;          </span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> p;          </span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>
<h3 id="25-K-个一组翻转链表"><a href="#25-K-个一组翻转链表" class="headerlink" title="25. K 个一组翻转链表"></a><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/reverse-nodes-in-k-group/">25. K 个一组翻转链表</a></h3><p>给你一个链表，每 <em>k</em> 个节点一组进行翻转，请你返回翻转后的链表。</p>
<p><em>k</em> 是一个正整数，它的值小于或等于链表的长度。</p>
<p>如果节点总数不是 <em>k</em> 的整数倍，那么请将最后剩余的节点保持原有顺序。</p>
<p><strong>进阶：</strong></p>
<ul>
<li>  你可以设计一个只使用常数额外空间的算法来解决此问题吗？</li>
<li>  <strong>你不能只是单纯的改变节点内部的值</strong>，而是需要实际进行节点交换。 </li>
</ul>
<p><strong>示例 1：</strong></p>
<p><img src="/myblog/2021/04/30/%E9%93%BE%E8%A1%A8%EF%BC%9A%E5%8F%8D%E8%BD%AC%E9%93%BE%E8%A1%A8/2.jpg" alt="img"></p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">输入：head &#x3D; [1,2,3,4,5], k &#x3D; 2</span><br><span class="line">输出：[2,1,4,3,5]</span><br></pre></td></tr></table></figure>

<p><strong>示例 2：</strong></p>
<p><img src="/myblog/%E5%8F%8D%E8%BD%AC%E9%93%BE%EF%BF%BD.jpg" alt="img"></p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">输入：head &#x3D; [1,2,3,4,5], k &#x3D; 3</span><br><span class="line">输出：[3,2,1,4,5]</span><br></pre></td></tr></table></figure>

<p><strong>示例 3：</strong></p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">输入：head &#x3D; [1,2,3,4,5], k &#x3D; 1</span><br><span class="line">输出：[1,2,3,4,5]</span><br></pre></td></tr></table></figure>

<p><strong>示例 4：</strong></p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">输入：head &#x3D; [1], k &#x3D; 1</span><br><span class="line">输出：[1]</span><br></pre></td></tr></table></figure>



<p><strong>提示：</strong></p>
<ul>
<li>  列表中节点的数量在范围 <code>sz</code> 内</li>
<li>  <code>1 &lt;= sz &lt;= 5000</code></li>
<li>  <code>0 &lt;= Node.val &lt;= 1000</code></li>
<li>  <code>1 &lt;= k &lt;= sz</code></li>
</ul>
<p>代码</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">/**</span></span><br><span class="line"><span class="comment"> * Definition for singly-linked list.</span></span><br><span class="line"><span class="comment"> * struct ListNode &#123;</span></span><br><span class="line"><span class="comment"> *     int val;</span></span><br><span class="line"><span class="comment"> *     ListNode *next;</span></span><br><span class="line"><span class="comment"> *     ListNode() : val(0), next(nullptr) &#123;&#125;</span></span><br><span class="line"><span class="comment"> *     ListNode(int x) : val(x), next(nullptr) &#123;&#125;</span></span><br><span class="line"><span class="comment"> *     ListNode(int x, ListNode *next) : val(x), next(next) &#123;&#125;</span></span><br><span class="line"><span class="comment"> * &#125;;</span></span><br><span class="line"><span class="comment"> */</span></span><br><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span></span><br><span class="line"><span class="class">&#123;</span></span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">ListNode* <span class="title">reverse</span><span class="params">(ListNode* a, ListNode* b)</span></span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        ListNode* pre = <span class="literal">nullptr</span>;</span><br><span class="line">        ListNode* cur = a;</span><br><span class="line">        ListNode* nxt = a;</span><br><span class="line">        <span class="keyword">while</span>(cur != b)</span><br><span class="line">        &#123;</span><br><span class="line">            nxt = cur-&gt;next;</span><br><span class="line">            cur-&gt;next = pre;</span><br><span class="line">            pre = cur;</span><br><span class="line">            cur = nxt;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> pre;</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">    <span class="function">ListNode* <span class="title">reverseKGroup</span><span class="params">(ListNode* head, <span class="keyword">int</span> k)</span> </span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="keyword">if</span>(!head)</span><br><span class="line">            <span class="keyword">return</span> <span class="literal">nullptr</span>;</span><br><span class="line">        </span><br><span class="line">        ListNode *a = head, *b = head;</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; k; ++i)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="keyword">if</span>(!b)</span><br><span class="line">                <span class="keyword">return</span> head;</span><br><span class="line">            b = b-&gt;next;</span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">        ListNode *newHead = <span class="built_in">reverse</span>(a, b);</span><br><span class="line"></span><br><span class="line">        a-&gt;next = <span class="built_in">reverseKGroup</span>(b, k);</span><br><span class="line">        <span class="keyword">return</span> newHead;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>


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